Adapted by Peter Au, George Brown College
McGraw-Hill Ryerson Copyright © 2011 McGraw-Hill Ryerson Limited.

9.1
z Tests about a Difference in Population Means:
One-Tailed Alternative
9.2
z Tests about a Difference in Population Means:
Two-Tailed Alternative
9.3
t Tests about a Difference in Population Means:
One-Tailed Alternative
9.4
t Tests about a Difference in Population Means:
Two-Tailed Alternative
9.5
z Tests about a Difference in Population
Proportions
9.6
F Tests about a Difference in Population
Variances
Copyright © 2011 McGraw-Hill Ryerson Limited 9-2

• Suppose a random sample has been taken from each of two different populations (populations 1 and 2) and suppose that the populations are independent of each other
• Then the random samples are independent of each other
• Then the sampling distribution of the difference in sample means is normally distributed or that each of the sample sizes n
1 and n
2 is large ((n
1
, n
2
) is at least 40) is more than sufficient
• We can easily test a hypothesis about the difference between the means
Copyright © 2011 McGraw-Hill Ryerson Limited 9-3

L01
• Suppose we wish to conduct a one-sided hypothesis test about μ
1
- μ
2
• The difference between these means can be represented by “D”
• i.e. μ
1
- μ
2
= D
• The null hypothesis is:
• H
0
: μ
1
- μ
2
= D
0
• The one-tailed alternative hypothesis is:
• H a
: μ
1
- μ
2
> D
0 or
• H a
: μ
1
- μ
2
< D
0
Copyright © 2011 McGraw-Hill Ryerson Limited 9-4

L01
• Often D
0 will be the number 0
• In such a case, the null hypothesis H
0
: μ
1
- μ
2
= 0 says there is no difference between the population means μ
1 and μ
2
• When D
0
= 0, each alternative hypothesis implies that the population means μ
1 and μ
2 differ
• Also note the standard deviation of the difference of means is:
 x
1
 x
2


1
2 n
1


2
2 n
2
Copyright © 2011 McGraw-Hill Ryerson Limited 9-5

L02
•
z

 x
1
 x
2


D
0

1
2 n
1

 2
2 n
2
•
•
Copyright © 2011 McGraw-Hill Ryerson Limited 9-6

• Reject H
0
: m
1
– m
2
= D
0 in favor of a particular alternative hypothesis at a level of significance if the appropriate rejection point rule holds or if the corresponding p-value is less than a
• Rules are on the next slide …
L01
Copyright © 2011 McGraw-Hill Ryerson Limited 9-7

Null Hypothesis: H
0
: m
1
– m
2
= D
0
Alternative Hypothesis Reject H
0 if:
H a
: μ
1
– μ
2
> D
0 z > z
α
H a
: μ
1
– μ
2
< D
0 z < -z
α
H a
: μ
1
– μ
2
≠ D
0
* |z| > z
α/2
*
* Note
For Two-Tailed Alternative either z > z a/2 or z < –z a/2
Copyright © 2011 McGraw-Hill Ryerson Limited p-value
Area under standard normal to the right of z
Area under standard normal to the left of –z
Twice the area under standard normal to the right of |z|
L01
L05
9-8

L02
• Test the claim that the new system reduces the mean waiting time
• Test at the a
= 0.05 significance level the null
• H
0
: m
1
– m
2
= 0 against the alternative H a
: m
1
– m
2
> 0
• Use the rejection rule H
0 if z > z a
• At the 5% significance level, z a
= z
0.05
= 1.645
• So reject H
0 if z > 1.645
• Use the sample and population data in Example
7.11 to calculate the test statistic z

 x
1
 x
2


1
2 n
1



2
2
D
0 n
2


8 .
79

5 .
14
 
0
4 .
7
100

1 .
9
100

3 .
65
0 .
2569

14 .
21
Copyright © 2011 McGraw-Hill Ryerson Limited 9-9

• Because z = 14.21 > z
0.05
= 1.645, reject H
0
• Conclude that m
1
– m
2 is greater than 0 and therefore it appears as though the new system does reduce the waiting time
• Alternatively we can use the p-value
• The p-value for this test is the area under the standard normal curve to the right of z = 14.21
• Since this p value is less than 0.001, we have extremely strong evidence that μ
1
- μ
2 is greater than 0 and, therefore, that the new system reduces the mean customer waiting time
L02
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-10

• The new system will be implemented only if it reduces mean waiting time by more than 3 minutes
• Set D
0
= 3, and try to reject the null H
0
: m
1 in favor of the alternative H a z

 x
1
 x
2

 n
1
2
1


 n
2
2
2
D
0


8 .
79

4 .
7
100
5 .
14


1 .
9
100

3
: m
1

– m
2
0 .
65
0 .
2569

> 3
2 .
53
– m
2
= 3
• z=2.53 > z
0.05
= 1.645, we reject H
0 in favor of H a
• There is evidence that the mean waiting time is reduced by more than 3 minutes
L02
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-11

• The p-value for this test is the area under the standard normal curve to the right of z = 2.53
• With Table A.3, the p-value is 0.5 – 0.4943 = 0.0057
• There is strong evidence against H
0
• Again there is evidence that the mean waiting time is reduced by more than 3 minutes
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-12

• A 95% confidence interval for the difference in the mean waiting time is: z




 x
1
 x
2
  z
0 .
025
 n
1
1
2

 n
2
2
2









8 .
79

5 .
14



3 .
3 .
65

15 , 4 .
0 .
5035

15


4 .
7
100

1 .
9
100


L02
Copyright © 2011 McGraw-Hill Ryerson Limited 9-13

Null Hypothesis: H
0
: m
1
– m
2
= D
0
Alternative Hypothesis Reject H
0 if:
H a
: μ
1
– μ
2
≠ D
0
* |z| > z
α/2
*
* Note
For Two-Tailed Alternative either z > z a/2 or z < –z a/2 p-value
Twice the area under standard normal to the right of |z|
L01
L05
Copyright © 2011 McGraw-Hill Ryerson Limited 9-14

• Provide evidence supporting the claim that the new system produces a different mean bank customer waiting time
• We will test H
0
: μ
1
- μ
2
= 0 versus H at the 0.05 level of significance a
: μ
1
= μ
2
≠ 0
• Reject H
0 than z
α/2
: μ
= z
1
- μ
2
0.025
= 0 if the value of |z| is greater
= 1.96
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-15

L03
L05
• Use the sample and population data in Example
7.11 to calculate the test statistic z

 x
1
 x
2


D
0

1
2


2
2 n
1 n
2


8 .
79

5 .
14
 
0
4 .
7
100

1 .
9
100

3 .
65
0 .
2569

14 .
21
• z = 14.21 is greater than z
0.025
= 1.96
• reject H
0
: μ
1
- μ
2
= 0 in favour of
• H a
: μ
1
= μ
2
≠ 0
• Conclude that μ
1
- μ
2 is not equal to 0
• There is a difference in the mean customer waiting times
Copyright © 2011 McGraw-Hill Ryerson Limited 9-16

• Testing the null hypothesis H
0
: μ
1 two conditions
1.
When variances are equal,

2.
When variances are unequal,
2
1

1

2
2
2
  
2
2
– μ
2
= D
0 under
L02
Copyright © 2011 McGraw-Hill Ryerson Limited 9-17

L02
L05
1.
When
 2
1
 
2
2
• The test statistic is: t

 x
1
 x
2


D
0 s 2 p


1 n
1

1 n
2


2.
When
 2
1
 
2
2
• The test statistic is:

X
1

X
2


D
0 t
 s
1
2
 s
2
2 n
1 n
2 df

 s
1
2
 n
1 s
1
2 n
1 n

1
1
2

  s
2
2 s
2
2 n
2 n
2
2
 n

2
1
2
 round down to the smallest whole number
9-18 Copyright © 2011 McGraw-Hill Ryerson Limited

L02
L05
• If sampled populations are both normal, but sample sizes and variances differ substantially, small-sample estimation and testing can be based on the following “unequal variance” procedure
Confidence Interval
 x
1
 x
2

 t a
/2 s
1
2 n
1
 s
2
2 n
2 t

Test Statistic
 x
1
 s
1
2 n
1 x
2

  s 2
2 n
2
D
0
For both the interval and test, the degrees of freedom are equal to df


 s
1
2 /n
1 s 2
1 n
1
/n

1
1
2

 s
2
2
 
/n
2

2 s 2
2 n
2
/n

2
1
2

Copyright © 2011 McGraw-Hill Ryerson Limited 9-19

H
0
: μ
1
– μ
2
= D
0
Alternative Reject H
0 if:
H a
: μ
1
– μ
2
> D
0
(One-Tailed) t > t
α
H a
: μ
1
– μ
2
< D
0
(One-Tailed) t < -t
α
H a
: μ
1
– μ
2
≠ D
0
(Two-Tailed) |t| > t
α/2
* where t
α
, t
α/2
, and p-values are based on (n
1
+ n degrees of freedom
2
- 2)
* either t > α a/2 or t < –t
α/2 p-value
Area under t distribution to the right of t
Area under t distribution to the left of t
Twice the area under t distribution to the right of |t|
L01
L05
Copyright © 2011 McGraw-Hill Ryerson Limited 9-20

• If the population of differences is normal, we can reject H
0
: m
D
= D
0 at the a level of significance
(probability of Type I error equal to a
) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than a
• We need a test statistic …
L02
Copyright © 2011 McGraw-Hill Ryerson Limited 9-21

L02
• The test statistic is:
D t= s
D
/

D
0 n
• D
0
= m
1
– m
2 is the claimed or actual difference between the population means
• D
0 varies depending on the situation
• Often D
0
= 0, and the null means that there is no difference between the population means
• The sampling distribution of this statistic is a t distribution with (n – 1) degrees of freedom
• Rules are on the next slide …
Copyright © 2011 McGraw-Hill Ryerson Limited 9-22

Alternative Reject H
0 if:
H a
: μ
D
> D
0
(One-Tailed) t > t
α p-value
Area under t distribution to the right of t
H
H a a
: μ
D
: μ
D
< D
0
≠ D
0
(One-Tailed) t < -t
α
(Two-Tailed) |t| > t
α/2
*
Area under t distribution to the left of t
Twice the area under t distribution to the right of
|t| where t
α
, t
α/2
, and p-values are based on (n – 1) degrees of freedom
* either t > α a/2 or t < –t
α/2
L01
L05
Copyright © 2011 McGraw-Hill Ryerson Limited 9-23

• Example 9.3 The Coffee Cup Case
• In order to compare the mean hourly yields obtained by using the
Java and Joe production methods, we will test H
0
: μ
1
μ
2
= 0 versus
H a
: μ
1
μ
2
> 0 at the 0.05 level of significance
• To perform the hypothesis test, we will use the sample information
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-24

• Unequal-variances procedure
• Consider the bank customer waiting time situation, recall that the bank manager wants to implement the new system only if it reduces the mean waiting time by more than three minutes
• Therefore, the manager will test the null hypothesis H hypothesis H
0 a
: μ
: μ
1
1
- μ
2
- μ
2
= 3
> 3 versus the alternative at α = 0.05
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-25

L02
L03
• Suppose
• n
1
= 100 and n
2
= 100, computing the sample mean and standard deviation of each sample gives x
1

8 .
79 s 2
1 x
2

4 .
8237

5 .
14 s 2
2

1 .
7927 t df


 s 2
1
 s 2
1 n
1

X
1

X
2
 s 2
1 n
1 n
1 n

1
1

2


 s
2
2 s 2
2   n
2 s 2
2
D
0 n
2 n
2


2 n

2
1
2



8 .
79

4 .

4 .
8237
8237
99
100
100

1 .
7927
 
1 .
7927
99
100

2
100

2
100

4 .
8237
5 .
14

 
1 .
7927
100
3

2 .
53

163 .
657

163
Copyright © 2011 McGraw-Hill Ryerson Limited 9-26

• t = 2.53 is greater than t
0.05
= 1.65
• Reject H
0
: μ
1
- μ
2
= 3 in favour of H a
:μ
1
2 μ
2
> 3 at α 0.05
• The new system reduces the mean customer waiting time by more than three minutes
• Examine the MegaStat output below
L02
L03
• t = 2.53, the associated p value is 0.0062, the very small p value tells us that we have very strong evidence against H
0
Copyright © 2011 McGraw-Hill Ryerson Limited 9-27

• Reject H
0
: μ
1
1.860
- μ
2
= 0 if t is greater than t
α
= t
0.05
=
• Test Statistic: t

 x
1
 x
2


D
0 s 2 p


1 n
1

1 n
2




811

750 .
2


0
435 .
1
1
5
1
5

4 .
6087
• t = 4.6087 > t
0.05
• We can reject H
0
= 1.860
• Conclude at α = 0.05 the mean hourly yields obtained by using the two production methods differ
• Note the small p-value in figure 9.1 indicates strong evidence against H
0
Copyright © 2011 McGraw-Hill Ryerson Limited
L02
L03
9-28

• Example 9.4 The Repair Cost Comparison Case
• Forest City Casualty currently contracts to have moderately damaged cars repaired at garage 2
• However, a local insurance agent suggests that garage 1 provides less expensive repair service that is of equal quality
• Forest City has decided to give some of its repair business to garage
1 only if it has very strong evidence that μ
1
, the mean repair cost estimate at garage 1, is smaller than μ
2
, the mean repair cost estimate at garage 2, that is, if μ
D
= μ
1
- μ
2 is less than zero
L02
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-29

• We will test H
0
: μ
D
= 0 (no difference) versus
H a
: μ
D
< 0 (difference – garage 1 costs are less than garage 2) at the 0.01 level of significance
• Reject if t < –t a
, that is , if t < –t
• With n – 1 = 6 degrees of freedom, t
0.01
0.01
= 3.143
• So reject H
0 if t < –3.143
L02
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-30

L02
L03
• Calculate the t statistic: t

D

D
0 s
D n


0 .
8
0 .
5033

0
7
 
4 .
2053
• Because t = –4.2053 is less than –t
0.01
= – 3.143, reject H
0
• Conclude at the a
= 0.01 significance level that it appears as though the mean repair cost at Garage 1 is less than the mean repair cost of Garage 2
• From a computer, for t = -4.2053, the p-value is 0.003
• Because this p-value is very small, there is very strong evidence that H less than m
2
0 should be rejected and that m
1 is actually
Copyright © 2011 McGraw-Hill Ryerson Limited 9-31

L02
L03
• Example 9.5 Coffee Cup Case (Revisited)
• In order to compare the mean hourly yields obtained by using the
Java and Joe methods
• Test H
0
: μ
1
- μ
2
= 0 versus H a
: μ
1
- μ
2
≠ 0 at α = 0.05
• Reject H
0 t
α/2
= t
0.025
: μ
1
- μ
2
= 2.306
= 0 if the absolute value of t is greater than
• df = n
1
+ n
2
- 2 = 5 + 5 - 2 = 8
• Test Statistic
 x
1
 x
2

 t
 s
2 p


1
 n
1 n
2
D
0
1




811

750 .
2


0
435 .
1
1
5
1
5

4 .
6087
• Because |t| = 4.6087 is greater than t
0.025
= 2.306, reject H
0 in favor of H a
• Conclude at 5% significance level that the mean hourly yields from the two catalysts do differ
Copyright © 2011 McGraw-Hill Ryerson Limited 9-32

• The p-value = 0.0017
• The very small p-value indicates that there is very strong evidence against H
0
(that the means are the same).
• Conclude on basis of p-value the same as before, that the two catalysts differ in their mean hourly yields
L02
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-33

L02
• The test statistic is: z=
 p
1

 p
2 p
1

 p
2

D
0
• D
0
= p
1
– p
2 is the claimed or actual difference between the population proportions
• D
0 is a number whose value varies depending on the situation
• Often D
0
= 0, and the null means that there is no difference between the population means
• The sampling distribution of this statistic is a standard normal distribution
Copyright © 2011 McGraw-Hill Ryerson Limited 9-34

• If the population of differences is normal, we can reject H
0
: p
1
– p
2
= D
0 at the a level of significance
(probability of Type I error equal to a
) if and only if the appropriate rejection point condition holds or, equivalently, if the corresponding p-value is less than a
• Rules are on the next slide …
L01
Copyright © 2011 McGraw-Hill Ryerson Limited 9-35

• For testing the difference of two population proportions
Alternative Reject H
0 if: p-value
H a
: p
1
– p
2
> D
0 z > z
α
H a
: p
1
– p
2
< D
0
H a
: p
1
– p
2
≠ D
0 z < -z
α
|z| > z
α/2
*
Area under the standard normal to the right of z
Area under the standard normal to the left of –z
Twice the area under the standard normal to the right of |z|
* either t > t a/2 or t < –t a/2
L01
L05
Copyright © 2011 McGraw-Hill Ryerson Limited 9-36

L02
L04
• If D
0
• If D
0

1
 ˆ
2 s p
1
 p
2
 s p
1
 p
2


1




1 n
1

1 n
2



1
 ˆ
2 p
1

1
 n
1 p
1

 p
2

1
 n
2 p
2

Copyright © 2011 McGraw-Hill Ryerson Limited 9-37

• Recall from example 7.15 that p
1 is the proportion of all consumers in the Toronto area who are aware of the new product and that p
2 is the proportion of all consumers in the Vancouver area who are aware of the new product
• To test for the equality of these proportions, we will test H
0
: p
1
- p
2
= 0 versus
0.05 level of significance
H a
: p
1
- p
2
≠ 0 at the
• Samples are large
L04
Copyright © 2011 McGraw-Hill Ryerson Limited 9-38

• Since H a
: p
1
- p
2
≠ 0 is of the form H a
: p
1
- p
2
≠ D0
• Reject H
0
: p
1
- p
2 greater than z
α/2
= 0 if the absolute value of z is
= z
0.05/2
= z
0.025
= 1.96
• 631 out of 1,000 randomly selected Toronto residents were aware of the product and 798 out of 1,000 randomly selected Vancouver residents were aware of the product, the estimate of p = p
1
= p
2 is

631

7982
1 , 000

1 , 000

1 , 429
2 , 000

0 .
7145
L04
Copyright © 2011 McGraw-Hill Ryerson Limited 9-39

• Test Statistic z

 p
1
 p
2


D
0

1




1 n
1

1 n
2




0 .
631

0 .
798
 
0
0 .
7145

0 .
2855



1
1 , 000

1
1 , 000


 
8 .
2673
• Because |z| - 8.2673 is greater than 1.96, we can reject H
0
: p
1
- p
2
= 0 in favour of H a
:p
1
- p
2
≠ 0
• The proportions of consumers who are aware of the product in
Toronto and Vancouver differ
• We estimate that the percentage of consumers who are aware of the product in Vancouver is 16.7 percentage points higher than the percentage of consumers who are aware of the product in Toronto
L02
L04
Copyright © 2011 McGraw-Hill Ryerson Limited 9-40

• The p value for this test is twice the area under the standard normal curve to the right of |z| = 8.2673
• The area under the standard normal curve to the right of 3.29 is
0.0005, the p-value for testing H
0 is less than 2(0.0005) = 0.001
• Extremely strong evidence that H
0
: p
1
- p
2
= 0 should be rejected
• Strong evidence that p
1 and p
2 differ
L03
Copyright © 2011 McGraw-Hill Ryerson Limited 9-41

L01
• Population 1 has variance
 variance

2
2
1
2 and population 2 has
• The null hypothesis, H
0 the same
, is that the variances are
• H
0
:

1
2 =

2
2
•
• That population has less variable, more consistent, measurements
• Suppose

1
2 >

2
2
• Let’s look at the ratios of the variances
• Test H
0
:

1
2 /

2
2 = 1 versus H a
:

1
2 /

2
2 > 1
Copyright © 2011 McGraw-Hill Ryerson Limited 9-42

• Reject H
0 in favor of H greater than 1 a if s
1
2 /s
2
2 is significantly
• s
1
2 is the variance of a random sample of size n from a population with variance

1
2
1
• s
2
2 is the variance of a random sample of size n from a population with variance

2
2
2
• To decide how large s
1
2 /s
2
2 must be to reject H
0 describe the sampling distribution of s
1
2 /s
2
2
,
• The sampling distribution of s
1 an F distribution
2 /s
2
2 is described by
Copyright © 2011 McGraw-Hill Ryerson Limited 9-43

• In order to use the F distribution
• Employ an F point, which is denoted F a
• F
A is the point on the horizontal axis under the curve of the F distribution that gives a right-hand tail area equal to α
• Shape depends on two parameters: the numerator number of degrees of freedom (df
1
) and the denominator number of degrees of freedom (df
2
)
Copyright © 2011 McGraw-Hill Ryerson Limited 9-44

• Suppose we randomly select independent samples from two normally distributed populations with variances

1
2 and

2
2
• If the null hypothesis H
0
:

1
2 /

2
2 = 1 is true, then the population of all possible values of s
1
2 /s
2
2 has an F distribution with df
1
= (n
1
– 1) numerator degrees of freedom and with df
2 denominator degrees of freedom
= (n
2
– 1)
L06
Copyright © 2011 McGraw-Hill Ryerson Limited 9-45

L06
•
a
a
• The value of F a depends on a
(the size of the right-hand tail area) and df
1 and df
2
• Different F tables for different values of a
• See:
• Table A.6 for a
= 0.10
• Table A.7 for a
= 0.05
• Table A.8 for a
= 0.025
• Table A.9 for a
= 0.01
Copyright © 2011 McGraw-Hill Ryerson Limited 9-46

• Independent samples from two normal populations
• Test H
0
:

1
2 =

2
2 versus H a
:

1
2 >

2
2
• Use the test statistic F = s
1
2 /s
2
2
• The p-value is the area to the right of this value of F under the F
• curve having df
1
(n
2
= (n
1
– 1) numerator degrees of freedom and df
– 1) denominator degrees of freedom
2
Reject H
0 at the a significance level if:
=
• F > F a
, or
• p-value < a
L06
Copyright © 2011 McGraw-Hill Ryerson Limited 9-47

• Independent samples from two normal populations
• Test H
0
:

1
2 =

2
2 versus H a
:

1
2 <

2
2
• Use the test statistic F = s
2
2 /s
1
2
• The p-value is the area to the right of this value of F under the F
• curve having df
1
(n
2
= (n
1
– 1) numerator degrees of freedom and df
– 1) denominator degrees of freedom
2
Reject H
0 at the a significance level if:
=
• F > F a
, or
• p-value < a
L06
Copyright © 2011 McGraw-Hill Ryerson Limited 9-48

L01
• Independent samples from two normal populations
• Test H
0
:

1
2 =

2
2 versus H a
:

1
2 ≠

2
2
• Use the test statistic F
 the larger of s
1
2 and s
2
2 the smaller of s
1
2 and s
2
2
• The p-value is twice the area to the right of this value
• of F under the F curve having df
1 degrees of freedom and df
2 degrees of freedom
= (n
2
Reject H
0
= (n
1
– 1) numerator
– 1) denominator at the a significance level if:
• F > F a/2
, or
• p-value < a
Copyright © 2011 McGraw-Hill Ryerson Limited 9-49

• The production supervisor wishes to use Figure
9.13 to determine whether σ
1
2 , the variance of the average production yields obtained by using the Java method, is smaller than σ
2
2 , the variance of the yields obtained by using the Joe method
• Test the hypotheses
• H
0
: σ
1
2 = σ
2
2 versus H a
: σ
1
2 < σ
2
2 or σ
1
2 > σ
2
2
L06
Copyright © 2011 McGraw-Hill Ryerson Limited 9-50

• Using the Excel output we can compute the test statistic
L06
F
 s
2
2 s
1
2

484 .
2
386

1 .
2544
Copyright © 2011 McGraw-Hill Ryerson Limited 9-51

• Compare this value with F a based on
• df
1
= n
2
- 1 = 5 - 1 = 4 numerator degrees of freedom and df
2
= n
1
- 1 = 5 - 1 = 4 denominator degrees of freedom at the 0.05 level of significance
• F
0.05
= 6.39
• F = 1.2544 is not greater than F
0.05
• we cannot reject H
0 at α = 0.05
= 6.39
• We cannot conclude that σ
1
2 is less than σ
2
2
L06
Copyright © 2011 McGraw-Hill Ryerson Limited 9-52

• It is possible to compare two populations using a one-tail or a two-tailed test
• Hypothesis tests can be conducted on such populations
(using CI’s, rejection points, or p-values)
• Populations may be independent or dependent (paired difference experiments)
• The value of σ may be known or unknown. This affects the type of test statistic we use (i.e. t or z)
• Independent tests can involve an equal variances assumption or an unequal variances assumption
• Two population variances can be compared using the F distribution
Copyright © 2011 McGraw-Hill Ryerson Limited 9-53

Copyright © 2011 McGraw-Hill Ryerson Limited 9-54

Copyright © 2011 McGraw-Hill Ryerson Limited 9-55

Copyright © 2011 McGraw-Hill Ryerson Limited 9-56

Copyright © 2011 McGraw-Hill Ryerson Limited 9-57